Home /
Expert Answers /
Algebra /
for-a-square-matrix-a-the-statement-a-lambda-i-v-0hasanon-zerosolutionv-is-equivalent-to-the-sta-pa336

For a square matrix

`A`

, the statement

` (A-lambda I)v=0hasanon-zerosolutionv `

is equivalent to the statement

`:'det(A-\lambda I)=0 : `

The quantity

`det(A-\lambda I)`

is a polynomial in

`\lambda `

known as the characteristic polynomial of

`A`

and its roots are the eigenvalues of

`A`

. This gives us a technique for finding the eigenvalues of

`A`

Let's check our understanding. i) The characteristic polynomial of

`A=([1,2],[2,1])`

is

`det(A-tI)=det([1-t,2],[2,1-t])=(1-t)^(2)-4,0 a `

Ordered

```
t_(1)=◻t_(2)=◻◻v_(1)=◻v_(2)=◻((1)/(2))(:1,2:)B=([5,-1,0],[-1,5,0],[0,0,-3])det(B-tI)=◻Bt_(1)=-3,t_(2)=◻,t_(3)=◻ 目, t_(1) this has roots (and hence B has eigenvalues)
t_(1)=-3,t_(2)=◻,t_(3)=◻ 目, t_(1) the roots of this polynomial are t_(1)= ◻ and t_(2)=
◻ Ordered the same way, the associated eigenvectors are
◻
v_(1)=
◻
and v_(2)= ◻
Note: the Maple notation for the vector ((1)/(2)) is (:1,2:)
ii
```